4 girls and 5 boys
(a) p(only boys) = \(\frac{^{5}C_{3}}{^{9}C_{3}}\)
= \(\frac{5!}{2!3!} \times \frac{3!6!}{9!}\)
= \(\frac{10 \times 1}{84}\)
= \(\frac{5}{42}\)
(b) p(more girls than boys) = p(3 girls and no boys) or p(2 girls and 1 boy)
Number of ways of selecting = \(^{4}C_{3} \times ^{5}C_{0} + ^{4}C_{2} \times ^{5}C_{1}\)
= \(4 \times 1 + 6 \times 5\)
= 34 ways.
Without restrictions, selection can be made in \(^{9}C_{3}\) ways.
= \(\frac{9!}{6! 3!}\)
= 84 ways.
\(\therefore\) p(more girls than boys) = \(\frac{34}{84}\)
= \(\frac{17}{42}\).